Consider $(\mathbb{Q}; \leq)$ and let $T$ be the theory of dense linear orderings without endpoints.
Let $\mathfrak{A}$ be the $\omega_1$ saturated model of $T$. Note that $|\mathfrak{A}|=2^{\aleph_0}$ since $\mathbb{R} \subset A$. However, $\mathbb{R} \neq A$ since $\mathfrak{A}$ has elements of the form "I am smaller than every positive rational, but larger than 0". Since $\mathbb{Q}$ is dense in $\mathbb{R}$, this element is not in $\mathbb{R}$. However, since $\mathfrak{A}$ is only $\omega_1$ saturated, we don't immediately get elements of the form "I am smaller than every positive real number, but larger than zero". For this to be the case, $\mathfrak{A}$ must be ${2^{\aleph_0}}^{+}$- saturated. Therefore, in some model which is ${2^{\aleph_0}}^+$- saturated, it is clear that $\mathbb{R}$ is not dense in that model.
Question: Is $\mathbb{R}$ dense in $\mathfrak{A}$ (the $\omega_1$ saturated model)?
If $\mathbb{R}$ were dense then so to would be $\mathbb{Q}$.